A rock is dropped from a height of 80 m and is in free fall. What is the velocity of the rock as it reaches the ground 4.0 seconds later?

The answer is 39.2 m/s

We need the velocity of the rock as it reaches the ground, and it is its final velocity. Since it is the free fall, we also must take into consideration gravitational acceleration (a). Therefore, the formula for the final velocity (v2) is:

v2 = v1 + at

v2 - the final velocity

v1 - the initial velocity

a - gravitational acceleration

t - time

As you can see, the distance in this case is not relevant.

It is known:

v1 = 0 m/s (since it was dropped)

a = 9.8 m/s²

t = 4 s

Therefore, the velocity of the rock as it reaches the ground (v2) is:

v2 = 0 + 9.8 · 4 = 39.2 m/s

Here is the solution below:

V = s / t

Where, S is the displacement and

t is the time taken.

Since displacement is expressed in meters and time taken in seconds. Velocity is expressed in meters/second or m/s.

V = 80m / 4.0 secs

V = 20 m/sec